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Q.∫x4−1x2+1 dx=\int \dfrac{x^4 - 1}{x^2 + 1}\,dx =

(a) x33+2x+k\dfrac{x^3}{3} + 2x + k
(b) x33−2x+k\dfrac{x^3}{3} - 2x + k
(c) x33+x+k\dfrac{x^3}{3} + x + k
(d) x33−x+k\dfrac{x^3}{3} - x + k
Bihar BsebBihar Board Intermediate 2021MCQ· 1mImportance★★★★★
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∫x4−1x2+1 dx=x33−x+k\displaystyle\int \dfrac{x^4 - 1}{x^2 + 1}\,dx = \dfrac{x^3}{3} - x + k.

Factor the numerator as a difference of squares: x4−1=(x2−1)(x2+1)x^4 - 1 = (x^2-1)(x^2+1). So

x4−1x2+1=(x2−1)(x2+1)x2+1=x2−1.\dfrac{x^4-1}{x^2+1} = \dfrac{(x^2-1)(x^2+1)}{x^2+1} = x^2 - 1.

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